Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie,
"ACE"
is a subsequence of"ABCDE"
while"AEC"
is not).Here is an example:
S ="rabbbit"
, T ="rabbit"
Return
3
.
//Dynamic programming //dp[i][j] stores the number of distinct subsequences of first j chars in T and first i chars in S //if(s[i - 1] == t[j - 1]) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; //else dp[i][j] = dp[i - 1][j]; //dp[0][0] = 1 //dp[0][1~m] = 0 //dp[1~n][0] = 1 //reference: http://www.cnblogs.com/ganganloveu/p/3836519.html class Solution { public: int numDistinct(string s, string t) { int n = s.size(); int m = t.size(); if(n == 0 && m != 0){ return 0; } if(n == 0 && m == 0){ return 1; } if(m == 0){ return 1; } int dp[n + 1][m + 1]; for(int i = 0; i <= n; i++){ dp[i][0] = 1; } for(int i = 0; i <= m; i++){ dp[0][i] = 0; } dp[0][0] = 1; for(int i = 1; i <= n; i++){ for(int j = 1; j <= m; j++){ if(s[i - 1] == t[j - 1]){ dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; } else{ dp[i][j] = dp[i - 1][j]; } } } return dp[n][m]; } };
public class Solution {
public int numDistinct(String s, String t) {
if(s == null || t == null || t.length() == 0) return 0;
int[] dp = new int[t.length()];
for(int i = 0; i<s.length(); i++){
char c = s.charAt(i);
for(int j=dp.length-1; j>=0; j–){
if(c == t.charAt(j)){
dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
}
}
}
return dp[t.length()-1];
}
}
URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html
Great! Your solution shrinks the space cost a lot.
I have a solution using less space:
public class Solution {
public int numDistinct(String s, String t) {
if(s == null || t == null || t.length() == 0) return 0;
int[] dp = new int[t.length()];
for(int i = 0; i
=0; j–){if(c == t.charAt(j)){
dp[j] = dp[j] + (j!=0?dp[j-1]: 1);
}
}
}
return dp[t.length()-1];
}
}
URL: http://traceformula.blogspot.com/2015/08/distinct-subsequences.html